Solution to Exercise 5.30.

(a) Table 5 shows the calculation of

$$E[XY] \; = \; 31 .$$

Thus, from the shortcut formula in the boxed proposition on p. 217, we obtain

$$Cov[X,Y] \; = \; 31 - 5.55*8.55 \; = \; -16.4525 .$$

x y p_XY x*y*p_XY

0.02 0.0

5 5 0.04 1.0

10 10 0.01 1.0

15 0.06 0.0

5 0.15 0.0

10 5 0.15 7.5

10 0.02 0.0

5 15 0.20 15.0

10 0.14 0.0

5 0.10 0.0

5 10 0.10 5.0

10 15 0.01 1.5

$$\cdot \cdot \cdot$$ 31.0

(b) We need to compute the standard deviations:

$$E[X^2] \; = \; (0^2)*0.20 + (5^2)*0.49 + (10^2)*0.31 \; = \; 43.25 .$$

$$\sigma_X \; = \; \sqrt{43.25-(5.55^2)} \; = \; 3.528101 .$$

$$E[Y^2] \; = \; (0^2)*0.07 + (5^2)*0.36 + (10^2)*0.36 + (15^2)*0.21 \; = \; 92.25 .$$

$$\sigma_Y \; = \; \sqrt{92.25-(8.55^2)} \; = \; 4.375786 .$$

Thus,

$$\rho_{XY} \; = \; \frac{-16.4525}{3.528101*4.375786} \; = \; -1.0657 .$$

Clearly some mistake here as correlations are always between -1 and 1.

OK, I see the problem. In part (a) I was on auto-pilot when I put in the y column in Table 5. The corrected version appears as Table 6. The correct value for the covariance is

$$Cov[X,Y] \; = \; 44.25 - 5.55*8.55 \; = \; -3.2025.$$

Thus,

$$\rho_{XY} \; = \; \frac{-3.2025}{3.528101*4.375786} \; = \; -0.2074398.$$

OK, that value isn't obviously wrong. I hope it is correct.

NOTE: It is kind of funny to get a negative correlation for the scores on two exams. That means a higher score on one exam tends to go with a lower score on the other.
 
 

x y p_XY x*y*p_XY

0.02 0.00

5 0.04 0.00

10 0.01 0.00

5 0.06 0.30

5 5 0.15 0.75

10 5 0.15 0.75

10 0.02 0.20

5 10 0.20 2.00

10 10 0.14 1.40

15 0.10 1.50

5 15 0.10 1.50

10 15 0.01 0.15

$$\cdot \cdot \cdot$$ 44.25

3/10/2001